Embedd SU(n) into an enlarged twisted Spin(2n) in terms of Lie groups precisely

Question

It is standard practice to show $$ SU(n) \subset SO(2n). $$

However, some post Is $SU(n) \subset \text{Spin}(2n)$? suggests that $\DeclareMathOperator\Spin{Spin}$ $$ SU(n) \subset \Spin(2n) (?). $$

I am very puzzled, because $1 \to \mathbb{Z}/2 \to \Spin(2n)\to SO(2n) \to 1$, so $SO(2n) \not \subset \Spin(2n) $. The $ SO(2n) $ is only a quotient group not a normal subgroup.

Is this answer $SU(n) \subset \Spin(2n)$ true of false, Is $SU(n) \subset \text{Spin}(2n)$?? I thought $SU(n) \not\subset \Spin(2n)$
However, is it possible to show that $$ SU(n) \times U(1) \subset \frac{\Spin(2n) \times \Spin(2)}{\mathbb{Z}/2}= \frac{\Spin(2n) \times U(1)}{\mathbb{Z}/2}? $$

$$ SU(n) \times SU(2) \subset \frac{\Spin(2n) \times \Spin(3)}{\mathbb{Z}/2}=\frac{\Spin(2n) \times SU(2)}{\mathbb{Z}/2}? $$ precisely?

score 3 · Accepted Answer · answered Apr 02 '20 at 04:35

3

Since $SU(n)$ is simply connected, any map $SU(n)\rightarrow SO(k)$ lifts to a map $SU(n)\rightarrow Spin(k)$. Further, there is a unique lift mapping the identity in $SU(n)$ to the identity in $Spin(k)$.

Apply this to the inclusion map $i:SU(n)\rightarrow SO(2n)$. Note that $i$ is injective, so any lift must also be injective, so there are always copies of $SU(n)$ in $Spin(2n)$

answered Apr 02 '20 at 04:35

Jason DeVito - on hiatus

51,998

Thanks voted up. I am puzzled, because 1→ℤ/2→(2)→(2)→1, so (2)⊄(2). The (2) is only a quotient group not a normal subgroup. Would this be a problem? – annie marie cœur Apr 02 '20 at 12:53
I don't see why it would be a problem. Suppose you replace $SU(n)$ with the trivial group. Would you say your exact sequences are ab problem in that case? – Jason DeVito - on hiatus Apr 02 '20 at 13:15
Before accept your answer, thanks I have a naive puzzle as bove, do you think that $\frac{Spin(n) \times Spin(10)}{\mathbb{Z}_2}$ contains $Spin(n) \times SU(5)$? – annie marie cœur Apr 02 '20 at 17:55
p.s. Your answer only says ${Spin(n) \times Spin(10)}$ contains $Spin(n) \times SU(5)$. – annie marie cœur Apr 02 '20 at 17:56
(2) what is the value of $k$? – annie marie cœur Apr 02 '20 at 20:51
may you know this math.stackexchange.com/q/3607058/141334 too ? – annie marie cœur Apr 02 '20 at 21:16
For (1), it may depend on which $\mathbb{Z}_2$ your referring to. For (2), $k$ is arbitrary, but for your application, it is $2n$. – Jason DeVito - on hiatus Apr 03 '20 at 02:43
For (1), I am wondering the common ℤ/2. Would that ()×(10)ℤ2 contain ()×(5)? – annie marie cœur Apr 03 '20 at 02:55
accepted, can you comment on this: math.stackexchange.com/q/3607058/141334 ? I believe it is true but I cannot show it in general... – annie marie cœur Apr 03 '20 at 19:44

Embedd SU(n) into an enlarged twisted Spin(2n) in terms of Lie groups precisely

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